One of Euler problems deals with reciprocal cycles. When you write unit fractions using their decimal representation, you soon find that 1/3 which reads 0.33333 can be written as 0.(3) to indicate that it has a 1-digit recurring cycle. Likewise, 1/6 is simply 0.1(6), and so on. Eric Hanchrow provided a very elegant solution to solve this problem in Racket, by using coprime and a for/fold form. That’s what I ended up using (with little rewrite) because I was struggling with this problem for days until I came across this solution. I made up for that cheating by doing some research, especially digging a bit to better understand the role of coprimes in this specific case.

#lang racket

(require math/number-theory)

(define (lrcdf q)
  (if (coprime? 10 q)
      (unit-group-order 10 q)
      0))

(for/fold ([p '(0 . 0)])
    ([x (in-range 2 1000)])
    (let ([l (lrcdf x)])
      (if (< (cdr p) l)
          (cons x l)
          p)))

As mentioned on the Racket docs, a set of integers is considered coprime if their greatest common divisor is 1. To put it in other words, any prime number that divides one does not divide the other.¹ Since we are interested in decimal expansion with a viniculum, this means that we need to work with integers coprime to 10.

The key point resolves around full reptend prime. According to Wikipedia, a proper prime is a prime p which ends in the digit 1 in base 10 and whose reciprocal in base 10 has a repetend (i.e., repeating decimal segment) with length p − 1. Each digit appears in the repetend the same number of times, (p-1)/10. A prime is a proper prime if and only if it is a full reptend prime and congruent to 1 mod 10, i.e. $10^k \equiv 1\ (\text{mod}\ p)$ for $k=p-1, k\nless p-1$. As a side note, we could remark that if this prime number is also safe, that is a prime of the form 2p + 1 where p is also a prime, then 1/p will produce a stream of p − 1 pseudo-random digits. Anyway, the idea is to find the first full reptend prime number below a limit, here $d < 1000$.

I also happened to find a very nice solution in Mathematica, by Nayuki:²

There’s more to see: The following fraction has 2997 recurring decimals!

$$ \frac{1}{998001} = 0.000001002003004005… $$

Indeed, you will successively find all 3-decimal numbers from 000 to 999, until it starts over again at 000. Check out James Grime’s explanations on Numberphile.